Answer
The solution for this system of equations is $\left(\frac{1}{2}, 2, -3\right)$.
Work Step by Step
First, let us see if we can modify any of the equations so that one of the variables in both equations is the same, differing only in sign.
We see that we can modify the second equation so that the $x$ terms of the first and second equations are the same but with opposite signs. We can multiply the second equation by $2$:
$(2)(-2x + 3y - z) = (2)(8)$
$-4x + 6y - 2z = 16$
Take the equation above and the first equation in the system to form the system
$ 4x - y + 2z = -6$
$-4x + 6y - 2z = 16$
Add the equations together:
$(4x-y+2z)+(-4x+6y-2z) = -6+16\\
5y=10$
Divide both sides by $5$ to solve for $y$
$y = 2$
We can substitute this value for $y$ in the third equation to find the value for $z$:
$2y+3z=-5\\
2(2) + 3z = -5$
$4 + 3z = -5$
Subtract $4$ from both sides of the equation to isolate the constants to the right side of the equation:
$3z = -9$
Divide each side by $3$ to solve for $z$:
$z = -3$
Now that we have the values of both $y$ and $z$, we can substitute them into one of the equations to find $x$. Let us use the second equation:
$-2x+3y-z=8\\
-2x + 3(2) - (-3) = 8$
$-2x + 6 + 3 = 8$
$-2x + 9 = 8$
Subtract $9$ from both sides to isolate constants to the right side of the equation:
$-2x = -1$
Divide both sides of the equation by $-2$ to solve for $x$:
$x = \frac{1}{2}$
The solution for this system of equations is $\left(\frac{1}{2}, 2, -3\right)$.
